Olm chào em, em nên viết đề bài bằng công thức toán học để thầy cô, bạn bè hiểu đúng yêu cầu đề bài, em nhé. Có như vậy mọi người mới trợ giúp em được tốt nhất. Cảm ơn em đã đồng hành cùng Olm.

Ta có: \(C=\left(-3\right)+\left(-3\right)^2+\left(-3\right)^3+\cdots+\left(-3\right)^{200}\)

=>\(\left(-3\right)C=\left(-3\right)^2+\left(-3\right)^3+\left(-3\right)^4+\cdots+\left(-3\right)^{201}\)

=>\(-3C-C=\left(-3\right)^2+\left(-3\right)^3+\cdots+\left(-3\right)^{201}-\left(-3\right)-\left(-3\right)^2-\cdots-\left(-3\right)^{200}\)

=>\(-4C=\left(-3\right)^{201}-\left(-3\right)=-3^{201}+3\)

=>\(4C=3^{201}-3\)

=>\(C=\frac{3^{201}-3}{4}\)

\(\left(13\frac{1}{4}-2\frac{5}{27}-10\frac{3}{6}\right)\)\(.230\frac{1}{25}+46\frac{3}{4}\)

\(=\)\(\left(\frac{53}{4}-\frac{59}{27}-\frac{21}{2}\right)\)\(.\frac{5751}{25}+\frac{187}{4}\)

\(=\frac{304803}{100}-\frac{12567}{25}-\frac{120771}{50}+\frac{187}{4}\)

\(=\frac{304803}{100}-\frac{50268}{100}-\frac{241542}{100}+\frac{4675}{100}\)

\(=\frac{304803-50268-241542+4675}{100}\)

\(=\frac{17668}{100}\)

\(=\frac{4417}{25}\)

f(2)=2\(^2\)+5
f(2)=9

cảm ơn cậu nhiều ạ

7,5 nha

7,5 ? chắc vậy

Ta có 2004/2003 =  2003+ 1/ 2003 = 1 + 1/2003

2003/2002= 2002 + 1/ 2002 = 1+ 1/2002

Do 1/2003 < 1/2002 =>  1 + 1/2003 <  1+ 1/2002 hay 2004/2003 < 2003/2002

\(\frac{2004}{2003}\)= 1,0004992

\(\frac{2003}{2002}\)= 1,0004995

Vậy ,\(\frac{2003}{2002}\)lớn hơn \(\frac{2004}{2003}\).

\(\frac{45^{10}.5^{20}}{75^{15}}\)

\(=\frac{5.3^{20}.5^{20}}{\left(5^2.3\right)^{15}}=\frac{5^{21}.3^{20}}{5^{30}.3^{15}}=\frac{5^{21}.3^{15}.3^5}{5^{21}.5^9.3^{15}}=\frac{3^5}{5^9}\).

\(\frac{45^{10}.5^{20}}{75^{15}}=\frac{\left(9.5\right)^{10}.5^{20}}{\left(3.25\right)^{15}}\)

                   \(=\frac{9^{10}.5^{10}.5^{20}}{3^{15}.25^{15}}\)

                   \(=\frac{\left(3^2\right)^{10}.5^{10}.5^{20}}{3^{15}.\left(5^2\right)^{15}}\)

                    \(=\frac{3^{20}.5^{30}}{3^{15}.5^{30}}\)

                    \(=\frac{3^{20}}{3^{15}}\) 

                     \(=3^5\)

                     \(=243\)

\(x:\frac{1}{3}=\frac{12}{99}:\frac{15}{90}\)

\(x:\frac{1}{3}=\frac{12}{99}:\frac{1}{6}\)

\(x:\frac{1}{3}=\frac{8}{11}\)

\(x=\frac{8}{11}X\frac{1}{3}\)

\(x=\frac{8}{33}\)

tk mk nhe

Ta có: x:1/3=12/99:15/90

x:1/3=8/11

x=8/33

Vậy x=8/33

\(\sqrt{2x+5}=11\Leftrightarrow2x+5=11^2\)

=> 2x + 5 = 121

=> 2x = 121 - 5 = 116

=> x = 58

\(\sqrt{2x+5}=11\)

\(\Leftrightarrow2x+5=11^2=121\)

\(\Leftrightarrow2x=121-5=116\)

\(\Leftrightarrow x=\frac{116}{2}=58\)